10041 - Vito's Family
Time limit: 3.000 seconds
Background
The world-known gangster Vito Deadstone is moving to New York. He has a very big family there, all of them living in Lamafia Avenue. Since he will visit all his relatives very often, he is trying to find a house close to them.
Problem
Vito wants to minimize the total distance to all of them and has blackmailed you to write a program that solves his problem.
Input
The input consists of several test cases. The first line contains the number of test cases.
For each test case you will be given the integer number of relatives r ( 0 < r < 500) and the street numbers (also integers)
where they live ( 0 < si < 30000 ). Note that several relatives could live in the same street number.
Output
For each test case your program must write the minimal sum of distances from the optimal Vito's house to each one of his relatives. The distance between two street numbers
s
i and s
j is d
ij= |s
i-s
j|.
Sample Input
22 2 4
3 2 4 6
Sample Output
24
排序求中位数即可。
优化：快速选择
排序代码：
复杂度：O(rlog r)
/*0.025s*/
#include<cstdio>
#include<algorithm>
using namespace std;
int s[505];
int main(void)
{
int t, r, sum;
scanf("%d", &t);
while (t--)
{
scanf("%d", &r);
for (int i = 0; i < r; ++i)
scanf("%d", &s[i]);
sort(s, s + r);
sum = 0;
for (int i = (r >> 1) + (r & 1); i < r; ++i)
sum += s[i] - s[r - i - 1];
printf("%d\n", sum);
}
return 0;
}
快速选择代码：
复杂度：O(r)，但系数大
速度反而比上面慢是因为这题r太小，系数对运行时间的影响大。
/*0.032s*/
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
int s[505];
int main(void)
{
int t, r, sum, mid;
scanf("%d", &t);
while (t--)
{
scanf("%d", &r);
for (int i = 0; i < r; ++i)
scanf("%d", &s[i]);
nth_element(s, s + (r >> 1), s + r);
mid = s[r >> 1], sum = 0;
for (int i = 0; i < r; ++i)
sum += abs(mid - s[i]);
printf("%d\n", sum);
}
return 0;
}